Write a quadratic function that fits the points (0, 6), (2, 4), and (3, 6).?

Answer 1

The general form of a quadratic function is:
f(x) = ax^2 + bx + c
you want all three points to satisfy this equation. Use each of your points by putting them into this equation to make a new equation.
For instance, the point (0,6) gives f(0) = c = 6
Do this with the other points to get 2 more equations. You'll have 3 equations and 3 unknowns (a, b, c).
Solve the system of equations to find a, b, and c. Then substitute these back into f(x) to get your function.

Answer 2

A quadratic function, in mathematics, is a polynomial function of the form y = ax^2 + bx + c, where a must be different to Zero.

Then we can write the following:

1. 6 = a*0^2 + b*0 + c
2. 4 = a*2^2 + b*2 + c
3. 3 = a*3^2 + b*3 + c

1. 6 = c
2. 4 = 4a + 2b + c
3. 3 = 9a + 3b + c

You solve the system to find a, b and c (already 6)

& Finally, you write these numbers in y = ax^2 + bx + c

Answer 3

Any quadratic function might properly be written as: f(x) = ax^2 + bx + c, the place a, b, and c are constants that count on what we choose the function to do. --- on account that (-a million, 8), (0, 6), and (a million, 2) lie on the graph of f(x), we see that: (i) f(-a million) = 8 ==> a - b + c = 8 (ii) f(0) = 6 ==> c = 6 (iii) f(a million) = 2 ==> a + b + c = 2. on account that c = 6, (i) and (iii) decrease to a - b = 2 and a + b = -4. including those provides: 2a = -2 ==> a = -a million. Then, substituting a = -a million returned into any equation provides b = -3. as a result, the required quadratic function is: f(x) = -x^2 - 3x + 6. i desire this permits!

Answer 4

Basic quadratic function has a formula: ax^2+bx+c=y

Here we have 3 points with coordinates: (0,6), (2,4), (3,6)

If we insert the values for x and y for each point, we get a simultaneous equation to help us solve for a, b and c.

i) a*0+b*0+c=6; c=6
II) a*4+b*2+6=4; 4a+2b=-2
iii) a*9+b*3+6=6; 9a+3b=0

(4a+2b=-2)*3
(9a+3b=0)*2

12a+6b=-6
18a+6b=0
------------------
-6a=-6
a=1

Insert the value of a in the original:
4a+2b=-2
4*1+2b=-2
b=-2-4/2= -6/2=-3

For a=1, b=-3, c=6 we can write quadratic equation for these values:

x^2-3x+6=y

Answer 5

y = ax^2 + bx + c

(0,6) --> 6 = c
(2,4) --> 4 = 4a + 2b + 6 --> 2a + b = -1
(3,6) --> 6 = 9a + 3b + 6 --> -3a - b = 0
--> a = 1 and b = -3 and c =6

y= x^2 - 3x + 6

Answer 6

when x = 0; y = 6
when x = 2; y = 4
when x = 3; y = 6
... when x = 0 or 3; y = 6
... x(x-3) = 6

x^2 -3x + 6 = 0

ANS: y = x^2 -3x +6

Answer 7

general form: y = ax^2 + bx +c

6 = a(0)^2 + b(0) + c
6 = c

a(2)^2 + b(2) + 6 = 4
4a + 2b = -2

a(3)^2 + b(3) + 6 = 6
9a + 3b = 0

4a + 2b = -2
4a + 2b +2 = 0

9a + 3b = 0

4a + 2b + 2 = 9a + 3b
b = -5a + 2

4 = a(2^2) + 2b + 6
4 = 4a + 2(-5a + 2) + 6
4 = 4a - 10a + 4 +6
-6 = -6a

a = 1

b = -5a + 2
b = -5(1) + 2
b = -3


y = x^2 - 3x + 6